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Given the function $f$ from $\Bbb R^+_0$ to $\Bbb R$ defined by the equation $$ f(x):=\begin{cases}\sin\Big(\frac 1 x\Big),\,\text{if }x\in\Bbb R^+\\0,\,\text{if }x=0\end{cases} $$ for any $x\in\Bbb R^+_0$ then we call Topologist sine curve the graph of $f$ that is the set $$ \mathcal G_f:=\biggl\{x\in\Bbb R^2:x_1\ge 0\wedge x_2=\sin\Big(\frac 1 {x_1}\Big)\biggl\}\cup\big\{\mathbf 0\big\} $$

Now it is a well know result that $\cal G_f$ is connected but not path-connected and thus I would like to discuss this about the path-disconnectedness of $\cal G_f$: so I do not really understand why it is possible I have to remove a part of $[0,1]$ and thus rescaling; then I do not understand why the path-disconnectedness of $\cal G_f$ follows by showing that $$ \sup\Big\{t\in[0,1]:f\big([0,t]\big)=\mathbf 0\Big\}=0 $$ finally I would like to see a formal proof about the subjectivity of $\sin\Big(\frac 1 x\Big)$ on $(0,\varepsilon)$ onto $[-1,1]$.

Please do not closed this question: unfortunately the users who wrote the linked answer and question are inactive by many time so that I cannot use comments to ask clarifications but the linked answer seems very simple and unfortuately I did not understand other proofs about the path-disconnectedness of $\cal G_f$ so that I really like to understand the linked proof: so could somoene help me, please?

Antonio Maria Di Mauro
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    For surjectivity of $\sin \frac 1x$ on $(0,\epsilon)$ onto $[-1,1]$, you just need to find two points whose images are $1$ and $-1$ (use appropriate values of $x$ for which the sine equals $\pm 1$) and apply the intermediate value theorem. – Sarvesh Ravichandran Iyer Jun 20 '22 at 11:14
  • @SarveshRavichandranIyer Okay, so I know that $\sin\Big(\frac 1 x\Big)=1$ if and only if $\frac 1 x=\frac \pi 2+2k\pi$ for any $k\in\Bbb Z$ . Now the last identity show that $x=\frac 1{\frac \pi 2+2k\pi}$ for any $k\in\Bbb Z$ but the sequence $$x_n:=\frac 1{\frac \pi 2+2n\pi}$$ converges to $0$ so that for any $\epsilon$ there exsist $n_0\in\Bbb N$ such that $$x_n\in (0,\epsilon)$$ which is such that $$\sin\Big(\frac 1{x_n}\Big)=1$$ Then by analogous argumens we can find $x\in (0,\epsilon)$ such that $\sin\Big(\frac 1 x\Big)=-1$, right? – Antonio Maria Di Mauro Jun 20 '22 at 11:22
  • However how explain other points? could you help me, please? – Antonio Maria Di Mauro Jun 20 '22 at 11:24
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    You're right, Antonio. (Some of the MathJax didn't render in your comment, but I read off the obvious expressions). Now, apply the intermediate value theorem as follows : imagine you got $z \in (0,\epsilon)$ such that $\sin \frac{1}{z} = 1$ and $z' \in (0,\epsilon)$ such that $\sin \frac{1}{z'} = -1$. Then, in the interval $[z,z']$ (or $[z',z]$ depending on which is bigger), for every $-1 \leq c \leq 1$ there must be a $y \in [z,z']$ such that $\sin \frac{1}{y} = c$. So what does this tell you about the range of $\sin \frac{1}{x}$ as a function on $(0,\epsilon)$? – Sarvesh Ravichandran Iyer Jun 20 '22 at 11:27
  • @SarveshRavichandranIyer Okay, so in this case we proved that $\sin\frac 1 x$ is surjective onto $[-1,1]$ in a subset of $(0,\epsilon)$ and thus it must surjective too in $(0,\epsilon)$ surely. So now how proceed with other doubts? – Antonio Maria Di Mauro Jun 20 '22 at 11:32
  • That's all I can do right now, unfortunately. I will leave the task of clarifying the other doubt to more capable posters. – Sarvesh Ravichandran Iyer Jun 20 '22 at 11:59
  • @SarveshRavichandranIyer Okay. Anyway thanks for your help. – Antonio Maria Di Mauro Jun 20 '22 at 12:21
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    Dear Antonio, your detailing made it easy for me to help you. I hope someone can give you an answer befitting your efforts. Thanks. – Sarvesh Ravichandran Iyer Jun 20 '22 at 12:32
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    Another proof of this property of the topologist’s sine curve, with two simpler examples presented first, is at https://kconrad.math.uconn.edu/blurbs/topology/connnotpathconn.pdf and you might prefer the argument there. – KCd Jun 20 '22 at 13:33

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I do not understand why the path-disconnectedness of $\mathcal{G}_f$ follows by showing that: $$\sup_{t\in[0,1]}\{f([0,t])=\{0,0\}\}=0$$

This is a misinterpretation of the linked answer. They don’t show that, indeed they suppose that by a “without loss of generality” argument.

Let me call the sine curve $T$ and the first and second components of a function into $T$ by subscript $1,2$ respectively. For such a function, $f_2=\sin(1/f_1)$ is forced (when $f_1\neq0$). Suppose $f$ is any path in $T$ joining $\{0,0\}$ to some other anonymous distinct point. It is impossible for the above supremum to equal $1$ (why?). Furthermore $f(0)=\{0,0\}$, and a point implicitly used later in their answer is that if $f_1(t)=0$ then $f_2(t)=0$ is forced. If $t_0\gt0$ is the supremum, by continuity $f(t_0)=\{0,0\}$. Then a rescaling of the function $f$ to the continuous path (with the same endpoints): $$g:[0,1]\to T, \,t\mapsto\begin{cases}f(t_0+t(1-t_0))&t\in(0,1]\\\{0,0\}&t=0\end{cases}$$Would set the supremum to $0$ and moreover not change the validity of the following argument.

The argument can be restated as follows. We take without loss of generality this modified path $g$ which has the supremum of $0$ and by continuity $g(t)\neq\{0,0\}$ for $t\gt0$. Continuously pullback the unit ball about the origin by $g$, yielding an interval $[0,t’)$ such that $|g_2|\lt1$ on this interval. This is the problematic step: try visualising this, it is impossible for any path touching $\{0,0\}$ to avoid $1$ continuously. Now by the intermediate value theorem $g_1$ attains an interval of nonzero values in $g_1[0,t’)$ (by the without-loss-of-generality argument and the italicised remark $g_1[0,t’)$ is a nondegenerate interval), so by the surjectivity of the reciprocal sine curve (consider $\frac{2}{n\pi},n\in\Bbb Z$, will eventually fall into any interval about $0$) it must be the case that $g_2([0,t’))=[-1,1]$, a contradiction (recall $g_2=\sin(1/g_1)$).

I hope that clarifies things, feel free to query further. Note that there is also nothing special about the number $1$ here; it is just aesthetic.

FShrike
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  • So if $f(1)\neq \mathbf 0$ then surely $t_0\neq 1$, right?; then if $f_1(t)=0$ it cannot be $f_2(t)\neq 0$ because otherwise it would be $$f_2(t)=\sin\frac 1{f_1(t)}$$ and this is impossible if $f_1(t)$ is zero, right?; then if $t<t_0$ there must exists $t<t^<t_0$ such that $f\big[[0,t^]\big]=\mathbf 0$ so that $f(t)=\mathbf 0$ for any $t<t_0$ and thus if $(t_n){n\in\Bbb N}$ is a sequence in $[0,t_0)$ converging to $t_0$ then by continuity $$f(t_0)=\lim{n\in\Bbb N}f(t_n)=\mathbf 0$$ but now I am not able to find a such sequence and this seem a problem to me, so what can you say about? – Antonio Maria Di Mauro Jun 20 '22 at 13:55
  • Now wait some minutes for other clarifications: I have to read carefully the last part of the answer. – Antonio Maria Di Mauro Jun 20 '22 at 14:03
  • $f^{-1}{0,0}$ is a closed set by continuity and contains its supremum. It cannot contain $1$ as $f(1)\neq{0,0}$ by definition of path, and thus $t_0\neq1$. For the same reason, $t_0\in f^{-1}{0,0}$, hence $f(t_0)={0,0}$. @AntonioMariaDiMauro It is often helpful to go for pure topological argument rather than using sequences, this clarifies the situation (context dependent - of course sequences are very common and useful in topology generally) – FShrike Jun 20 '22 at 14:08
  • I should add that by definition of the topologist’s sine curve, $f_1(t)=0$ forces $f_2(t)=0$ (as already remarked, and as used later on in the post) – FShrike Jun 20 '22 at 14:09
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    The “same reason” is: ${0,0}$ is closed and $f$ is continuous, hence $f^{-1}{0,0}$ is closed and contains its limit points, in particular its supremum. As $t_0=\sup f^{-1}{0,0}$, $t_0\in{0,0}$ and $f(t_0)={0,0}$. We have $t_0\neq1$ immediately from this since $f$ must connect to another point distinct from the origin. – FShrike Jun 20 '22 at 14:14
  • Oh, yeah!!! So now we let go forward to the second part. So if $f$ is continuous at $0$ then there exists $\delta\in\Bbb R^+$ such that $$|g_2(t)|=|g_2(t)-0|=|g_2(t)-g(0)|<1$$ for any $t\in[0,\delta)$, right? Now if $0$ is equal to $t_0$ then any $t\in(0,1]$ is not in ${t\in[0,1]:g\big[[0,t]\big]}$ and thus in particular for any $\delta'<\delta$ there exsits $t'\in[0,\delta']$ such that $g_1(t)\neq 0$ so that, as you stated, we can assume that [0,g'(t)) is non degenerate, right? – Antonio Maria Di Mauro Jun 20 '22 at 14:17
  • @AntonioMariaDiMauro Calling the open unit ball about the origin $B$, I prefer to just consider that $g^{-1}(B)$ is open and must contain an open interval about $0$, $[0,t’)$. As remarked $g_1(t)=0\implies g_2(t)=0,g(t)={0,0}$ but the linked answer (and myself) chose $g$ so that only $t=0$ satisfies $g(t)=0$. So, $g_1(t’)\neq0$. – FShrike Jun 20 '22 at 14:20
  • Okay, now it seem all clear so I up voted the answer but give me some minutes to accept it: I would like to rewrite it in my notes so that if then I will not have doubt then I surely accept it otherwise I will ask you other clarifications. Anyway thanks very much for your additional explanations: you was really kind so thanks really. – Antonio Maria Di Mauro Jun 20 '22 at 14:30
  • @AntonioMariaDiMauro You’re very welcome! Aim to take the notes in a compact way as I have done; no need for sequences, epsilons, deltas,... (replace the open unit ball about the origin with whatever open neighbourhood you like, so long as the $y$-value is never $1$ or $-1$, then $|g_2|\lt1$ on $[0,t’)$) – FShrike Jun 20 '22 at 14:34
  • Hi, just now I am rewriting the proof in my notes: I was thinking that $f^{-1}{0,0}$ is different from $$F_O:=\big{t\in [0,1]:f\big[[0,t]\big]\big}$$ but it contains surely it, right? Now as you observe $f^{-1}{0,0}$ is closed and so it contains all limit points and thus also it supremum that is also an upper bound for $F_O$: so observing that $1$ is not element of $f^{-1}{0,0}$ we conclude that $1$ is different from $t_0$, right? – Antonio Maria Di Mauro Jun 21 '22 at 07:54
  • @AntonioMariaDiMauro You have not finished your definition of $F_O$ - such that $f[[0,t]]$ what? But yes, $t_0\lt1$ is forced (because $f(1)$ has to be a nonzero point) and the path $f$ always induces such a modified path $g$, and $g$ arrives at the contradiction already explained (therefore $f$ arrives at this contradiction too) – FShrike Jun 21 '22 at 08:02
  • After all, if $f$ leaves $0$ and then loops back to $0$, and then leaves $0$ to arrive at the other distinct endpoint, the beginning loop is, well, irrelevant to the proof – FShrike Jun 21 '22 at 08:05
  • Sorry, I wanted define $$F_O:=\big{t\in [0,1]:f\big[[0,t]\big]=\mathbf 0\big}$$ so that I observed that generally only the strictly inclusion $$F_O\subset f^{-1}{0,0}$$ and thus I concluded that generally $F_O$ is not closed so that I do not understand why $t_0$ lies in $F_O$, that's all. Anyway if $\alpha$ is the sup of $f^{-1}{0,0}$ then surely $\alpha$ lies in $f^{-1}{0,0}$ and thus $\alpha$ is different from $1$: so if $F_O$ is contained in $f^{-1}{0,0}$ then $t_0$ is less than $\alpha$ and so different from $1$, right? – Antonio Maria Di Mauro Jun 21 '22 at 08:15
  • So could you explain why $t_0$ lies in $F_O$ if this is not generally closed, please? It seem I have to use sequences to prove it, that is I have to use sequences to prove that $f(t_0)$ is $\mathbf 0$ – Antonio Maria Di Mauro Jun 21 '22 at 08:15
  • Well, $f(t_0)={0,0}$ still holds, so the modified path $g$ is continuous, and this is all that matters! In fact it is cleaner to forget about $F_O$ entirely and just let $t_0=\sup{f^{-1}{0,0}|}$. The proof works the same and all your concerns can be left aside. @AntonioMariaDiMauro But working with $F_O$ is ok, since $g_1[0,t’)$ still contains a nondegenerate interval about $0$, and this is all we need (I should have written $g_1[0,t’)$ rather than $[0,g_1(t’))$, but the latter is ok to write if we work with $f^{-1}{0,0}$ instead). I think $t_0$ in $F_0$ is true, but it is unimportant. – FShrike Jun 21 '22 at 08:19
  • $f(t_0)={0,0}$ is true because: every neighbourhood of $f(t_0)$ continuously pulls back to a neighbourhood of $t_0$; by definition, every neighbourhood of $t_0$ contains points which $f$ maps to ${0,0}$, hence every neighbourhood of $f(t_0)$ contains ${0,0}$. Since the sine curve is a Hausdorff space, this forces $f(t_0)={0,0}$. That furthermore also forces $t_0\in F_0$ I think; for all $0\le t\le t_0$, can you see why $f(t)={0,0}$? @AntonioMariaDiMauro – FShrike Jun 21 '22 at 08:39
  • Okay, so finally we know that there exists $\delta\in\Bbb R^+$ such that $|g_2(t)|<1$ for any $t\in[0,\delta)$. Now the supremum of $F_O$ is $0$ and if $\delta$ is positive then there exist $t_\delta\in[0,\delta)$ such that $g_1(t_\delta)$ is positive because otherwise $g_2(t)$ would be $0$ for any $t\in[0,\delta)$ and this would mean that any $t\in[0,\delta)$ is an element of $F_O$ and this is impossible. So we conclude that $\big[0,g_1(t_\delta)\big)$ is a non-degenerate interval. – Antonio Maria Di Mauro Jun 21 '22 at 10:23
  • Now by the intermediate value theorem for any $\gamma\in\big(0,g_1(t_\delta)\big)$ there exists $t\in(0,t_\delta)$ such that $g_1(t)=\gamma$; however $\sin\frac 1 x$ is surjective in $\big(0,g_1(t_\delta)\big)$ and thus there exists $\gamma_1\in\big(0,g_1(t_\delta)\big)$ such that $$\sin\frac 1 {\gamma_1}=1$$ so that finally there exists $t_1\in(0,t_\delta)$ such that $$g_2(t)=\sin\frac 1{g_1(t_1)}=\sin\frac 1{\gamma_1}=1$$ which is a contradiction. So is this correct? – Antonio Maria Di Mauro Jun 21 '22 at 10:23
  • @AntonioMariaDiMauro It’s correct. If you chose to work without $F_0$ (so, set $t_0=\sup f^{-1}{0,0}$) you could simplify the language a bit. But all’s well – FShrike Jun 21 '22 at 10:37
  • @Okay, now it is all clear: forgive my bad English and forgive if I do not immediately understand your statements but unfortunately I have some difficults with not latin words and not latin grammar constructions so that I think I will to wrote an answer clarification above. Anyway thanks very much for your assistance: you were really helpful and thus I approved your answer now. See you soon. :) – Antonio Maria Di Mauro Jun 21 '22 at 10:41
  • @AntonioMariaDiMauro You’re welcome. Don’t worry about your language, you speak English fine enough - despite doing well at languages in school, I wouldn’t be able to write an MSE post in, eg, French... – FShrike Jun 21 '22 at 10:43